Let y=f(x) be the curve satisfy the differential equation 2xydy=(x2+y2+1)dx such that f(1)=0. Then the value of f2(2) is
A
3
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B
3.00
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C
03
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D
3.0
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Solution
Given: 2xydy=(x2+y2+1)dx ⇒2ydydx−y2x=x+1x
Let y2=t⇒2ydydx=dtdx ∴dtdx−tx=x+1x I.F.=1x
Now, the general solution: t⋅1x=∫1x(x+1x)dx ⇒y2x=x−1x+C
Since f(1)=0, therefore C=0 ∴y2=x2−1
For x=2: f2(2)=y2(2)=22−1=3